consider a regular \(n-gon\). show that a "flip", followed by a "ccw rotation" equals the "cw rotation" followed by a "flip": \[\large rf=fr^{-1}\]

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consider a regular \(n-gon\). show that a "flip", followed by a "ccw rotation" equals the "cw rotation" followed by a "flip": \[\large rf=fr^{-1}\]

Mathematics
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That's a concrete example, to prove it abstractly ...
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so we want to show r * f = f * r^-1 , where * stands for composition we can use the fact that f*f = Identity
i think so :) \(f = f^{-1}\) is kinda obvious as reflection undoes itself..
https://en.wikipedia.org/wiki/Dihedral_group#Group_structure
r * f = r *f^-1 = (r^-1)^-1 * f^-1 = ( f * r^-1 )^-1
Apparently I proved that r*f and f*r^-1 are inverses, but that wasn't the question :)
using (a*b)^-1 = b^-1 * a^-1
Haha nice :) that is also clear from shoe socks principle \((rf)^{-1} = f^{-1}r^{-1} = fr^{-1}\)
is there a proof r f = f r^-1 without having to go into the geometry , considering when n is even and odd, etc
not so sure, i got stuck on this while attempting below exercises https://i.gyazo.com/6b1fbd9dc197477edffe9987ca355527.png
they are from Gallian abstract algebra book
You can use the fact that frf = r^-1
http://prntscr.com/899tqo
given f*f = I (f * r* f) = r^-1 r* f = I * (r*f) = (f* f )* r * f = f * ( f* r * f ) = f * r ^-1
'given' meaning these are theorems for rigid transformations of n-gon
f r f = r^-1 In geometric terms: in the mirror a rotation looks like an inverse rotation.
that looks interesting!
but aren't we trying to prove the same thing
\[\large rf=fr^{-1}\] left multiplying both sides by \(f\) gives \[\large frf=r^{-1}\]
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